I have the formular $\frac{d(y(t)*a)}{d(t*b)}$ with a and b being constants, which can be changed to $\frac{a}{b}\frac{d(y(t))}{d(t)}$.
My question is why this is possible?
For the numerator it's clear because when deriving a formular the constants don't change, but why can this rule also be applied for the denominator in a derivative?
Write $w=y(t)a$ and $z=tb$. Then $$\frac{dw}{dt}=a\frac{d(y(t))}{dt}.$$ Since $t=\frac{z}{b}$, we get $$\frac{dt}{dz}=\frac{1}{b}.$$ Hence, $$\frac{d(y(t)a)}{d(tb)}=\overbrace{\frac{dw}{dz}=\frac{dw}{dt}\cdot\frac{dt}{dz}}^{\text{Chain Rule}}=a \frac{d(y(t))}{dt}\cdot\frac{1}{b}=\frac{a}{b}\cdot\frac{d(y(t))}{dt}.$$
Apply the chain rule.
$\frac{dy(t)}{dt} = \frac{dy(t)}{dbt}*\frac{dbt}{dt}=b\frac{dy(t)}{dbt}$
So $\frac{dy(t)}{dbt} = \frac 1b\frac{dy(t)}{dt}$
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Also Note: Let $w(t) = y(t/b)$ then $y(t) = y(\frac {bt}b) = w(bt)$
So $\frac {dy(t)}{dbt}=\frac{dw(bt)}{dbt} = \lim_{h\rightarrow 0}\frac{w(bt + h) - w(bt)}{h} = \lim_{h\rightarrow 0}\frac{y(\frac{bt + h}b) - y(t)}{h}$
$=\lim_{h\rightarrow 0}\frac{y(t + h/b) - y(t)}{h}=\lim_{j =h/b\rightarrow 0}\frac{y(t + j) - y(t)}{bj}=\frac 1b\lim_{j \rightarrow 0}\frac{y(t + j) - y(t)}{j}$
$= \frac 1b*\frac{dy(t)}{dt}$
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A third way using the other notation: $f'(x) = df(x)/dx$ so $df(bx)/dbx = f'(bx)$. Let $w(t) = y(t/b)$ so $w'(t) = \frac 1b y'(t/b)$. So as $y(t) = w(bt)$ and $w'(bt) = \frac 1b y'(bt/b) = \frac 1b y'(t)= \frac 1b \frac{dy(t)}{dt}$ but as well $w'(bt) = \frac{dw(bt)}{dbt} = \frac{dy(t)}{dbt}$.
Because $d(b t) = b~d(t)$ for constant $b$.